Forum:How do you represent the LVO ?/Is There a Non-OCF Notation Surpassing the LVO ?
At 8:50 in the below video the person explaining \psi(\Omega^{\Omega^{\Omega}}) stated it is the supremum of Veblen notation, but not necessarily \varphi(1,0,0,0,\ldots) , but instead was Veblen notation nested within each other infinitely. Unfortunately he never explained how this could be done, so I took an educated guess. First, take \varphi(1,0,0,0,\ldots) and represent it by a v_0 , next make the rule that v_{n+1}=\varphi(v_n,0,0,0,\ldots) . I feel the Large Veblen Ordinal is probably equal to the first fixed point of \xi\mapsto v_{\xi} ; if I am mistaken please correct me and explain a procedure involving a nesting of \varphi(1,0,0,0,\ldots) that can create the Large Veblen Ordinal. In addition, is there an ordinal notation that can surpass \psi(\Omega^{\Omega^{\Omega}}) without resorting to infinite collapsing functions, and which reaches higher ordinals such as \psi(\Omega^{\Omega^{\Omega^{\Omega}}}) , or \psi(\psi_I(0)) ? Transfinitary-argument Veblen functions vex me, and so I want to find a simpler alternative. Comments are appreciated.Edwin Shade (talk) 03:05, November 27, 2017 (UTC) : Well, there's the pair-sequence notation which follows a pretty intuitive lexicographic ordering and gets you up to ψ(Ωω). It's technically a notation for large numbers rather than ordinals, but it can easily be used as a notation for ordinals by snipping the final "n" in the expression. In this notation, the LVO will be written as: : LVO = (0,0)(1,1)(2,1)(3,1)(4,1) : You can also use nested trees of ordinals to reach ψ(ψɪ(0)) but these are incredibly confusing so they probably defeat the entire purpose of your question. : As for Transfinitary-argument Veblen functions, they are actually easy to understand. Instead of writing things like φ(1,0,0,0,...) (which get confusing very quickly), it is better to write the positions of the arguments like this: : φ(2,0,0,7,3) = φ( 2@4, 0@3, 0@2, 7@1, 3@0) (read as "phi of 2 at 4, 0 at 3, 0 at 2, 7 at 1, 3 at 0") : The key here is that we're allowed to omit the zeros, and that even in the Tranfinitary-argument Veblen functions we can only have a finite number of nonzero arguments. : So we can write: : φ(2,0,0,7,3) = φ( 2@4, 7@1, 3@0) : And what you wrote as φ(1,0,0,0,...) can be written as: : φ(1,0,0,0,0, ...) = φ( 1 @ ω ). : And this, really, works in exactly the same way as ordinary Veblen functions. So just like we have (say): The first fixed point of ξ→φ(ξ,0,0,0) is φ(1,0,0,0,0) : We also have: The first fixed point of ξ→φ(ξ @ ω ) is φ(1 @ ω+1 ) : This is the ordinal you've mentioned in your question, and it is far smaller than the LVO. Remember that you can have any ordinal as a "position", so we can have φ(1 @ ε₀ ) or φ(1 @ Γ₀ )... The limit of this system would be: : LVO = The first fixed point of ξ→φ( 1 @ ξ ) : You can actually extend this further by starting a new Veblen-like hierarchy with the seperator "@". Gotta be careful about how you do that, but this can get you up to the BHO at the very least. PsiCubed2 (talk) 09:08, November 27, 2017 (UTC) : Thank you PsiCubed2, I think I understand the LVO now. Edwin Shade (talk) 01:00, November 28, 2017 (UTC) : What psi said. Regarding to your request for a way to represent LVO using phi style notation instead of ocf, check this article, cantorsattic . Also, one last thing, i remember from a while ago that that video uses some non-standard (perhaps wrong) stuff beyond bachmann howard ordinal so beware. (already up way too late to look up the exact specifics) : Chronolegends (talk) 09:50, November 27, 2017 (UTC) :: I second Chrono's warning regarding those videos. If I remember correctly, some serious errors started to creep in well below the BHO level. PsiCubed2 (talk) 14:36, November 27, 2017 (UTC)